Question

In given Fiig., two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) $\Delta \text{PAC~}\Delta \text{PDB}$
(ii) PA . PB = PC . PD

Answer 1

(i)
Step 1: Finding exterior angle
$\text{In}\Delta \text{PAC and}\Delta \text{PDB,}$
$\angle \text{PAC}=\angle \text{PDB}$ ...(Exterior Angle is equal to opposite interior angle in cyclic quadrilateral)

Step 2: Applying similarity in $\Delta \text{PAC and}\Delta \text{PDB}$

$\Rightarrow \text{In}\Delta \text{PAC and}\Delta \text{PDB,}$
$\angle \text{PAC}=\angle \text{PDB}$ ...(Exterior angle is equal to opposite interir angle in cyclic quadrilateral)
$\angle \text{APC}=\angle \text{BPD}$ (common in both triangles)
By AA-criiterion of similarity,
$\Delta \text{PAC~}\Delta \text{PDB}$
Hence proved.

(ii)
Step 1: Applying A-A similarity
In $\Delta \text{PAC and}\Delta \text{PDB}$

$\Rightarrow \text{In}\Delta \text{PAC and}\Delta \text{PDB,}$
$\angle \text{PAC}=\angle \text{PDB}$ ...(Exterior angle is equal to opposite interior angle in cyclic quadrilateral)
$\angle \text{APC}=\angle \text{BPD}$ (common in both triangles)
By AA-criterion of similarity,
$\Delta \text{PAC~}\Delta \text{PDB}$
Hence proved.

Step 2: Using property of similar triangles
Sides are proportional in similar triangles
$\frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}=\frac{\text{CA}}{\text{BD}}\Rightarrow \frac{\text{PA}}{\text{BD}}=\frac{\text{PC}}{\text{PB}}$

$\Rightarrow \text{PA}\times \text{PB}=\text{PC}\times \text{PD}$

Hence proved.